結果

問題 No.2738 CPC To F
ユーザー lam6er
提出日時 2025-04-16 00:00:37
言語 PyPy3
(7.3.15)
結果
WA  
実行時間 -
コード長 1,536 bytes
コンパイル時間 380 ms
コンパイル使用メモリ 81,704 KB
実行使用メモリ 89,020 KB
最終ジャッジ日時 2025-04-16 00:02:46
合計ジャッジ時間 2,715 ms
ジャッジサーバーID
（参考情報） 		judge5 / judge2
このコードへのチャレンジ
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ファイルパターン 結果
sample AC * 2
other AC * 5 WA * 17
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ソースコード

diff # 

n = int(input())
s = input()

# Step 1: Count existing CPCTF substrings
existing_list = []
for i in range(n - 4):
    if s[i] == 'C' and s[i+1] == 'P' and s[i+2] == 'C' and s[i+3] == 'T' and s[i+4] == 'F':
        existing_list.append(i)

# Step 2: Find all CPC that are preceded by CPCT and do not overlap with existing CPCTF's first 3 characters
candidates = []
for i in range(n - 6):
    # Check if the current position starts with CPCT
    if s[i] == 'C' and s[i+1] == 'P' and s[i+2] == 'C' and s[i+3] == 'T':
        # Check if followed by CPC
        cpc_start = i + 4
        if cpc_start + 2 < n and s[cpc_start] == 'C' and s[cpc_start+1] == 'P' and s[cpc_start+2] == 'C':
            # Check overlap with existing CPCTF's first 3 characters
            overlap = False
            for j in existing_list:
                existing_start = j
                existing_end = j + 2
                # CPC spans from cpc_start to cpc_start + 2
                if not (cpc_start + 2 < existing_start or cpc_start > existing_end):
                    overlap = True
                    break
            if not overlap:
                candidates.append((cpc_start + 2, cpc_start))  # (end, start)

# Step 3: Greedily select non-overlapping CPC conversions
candidates.sort()
selected = []
last_end = -1
for end, start in candidates:
    if start > last_end:
        selected.append((start, end))
        last_end = end

# The answer is the existing count plus the number of selected conversions
print(len(existing_list) + len(selected))
0